If $x+y=\frac{\pi}{2}$, then the maximum value of $\sin x$.siny is

If $x+y=\frac{\pi}{2}$, then the maximum value of $\sin x$.siny is
  1. $\frac{1}{2}$
  2. $\frac{-1}{2}$
  3. $\frac{-1}{\sqrt{2}}$
  4. $\frac{1}{\sqrt{2}}$

Solution

$x+y=\frac{\pi}{2} \quad \Rightarrow \quad y=\frac{\pi}{2}-x$ $\begin{array}{l} \sin x \cdot \sin y=\sin x \cdot \sin \left(\frac{\pi}{2}-x\right)=\sin x \cos x=\frac{2 \sin x \cdot \cos x}{2}=\frac{\sin 2 x}{2} \\ \text { We know, }-1 \leq \sin 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq \frac{\sin 2 x}{2} \leq \frac{1}{2} \end{array}$ So maximum value is $\frac{1}{2}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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